Construction of Piecewise Linear Wavelets

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Construction of Piecewise Linear Wavelets.Jiansheng CaoEast Tennessee State University

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Construction of Piecewise Linear Wavelets

A Thesis

Presented to the Faculty of the Department of Mathematics

East Tennessee State University

In Partial Fulfillment

of the Requirements for the Degree

Master of Science in Mathematical Sciences

by

Jiansheng Cao

August 2002

Don Hong, Chair

Robert Gardner

Jeff Knisley

Jay Boland

Keywords: Piecewise, Linear Wavelets, Smaller Support

ABSTRACT

Construction of Piecewise Linear Wavelets

Jiansheng Cao

It is well known that in many areas of computational mathematics, wavelet basedalgorithms are becoming popular for modeling and analyzing data and for providing

efficient means for hierarchical data decomposition of function spaces into mutu-ally orthogonal wavelet spaces. Wavelet construction in more than one-dimensional

setting is a very challenging and important research topic. In this thesis, we firstintroduce the method of constructing wavelets by using semi-wavelets. Second, we

construct piecewise linear wavelets with smaller support over type-2 triangulations.Then, parameterized wavelets are constructed using the orthogonality conditions.

2

Copyright by Jiansheng Cao 2002

3

DEDICATION

This thesis is dedicated to Lianyun Song, my wife, and XiangKun Cao, my son,

who have supported my efforts to complete my graduate degree. Thanks for all your

love and support.

4

ACKNOWLEDGEMENTS

A special thanks to my thesis advisor, Dr. Don Hong, who has been patient with

me through the entire process. And a word of thanks to the rest of my committee

who has graciously given their time to support my thesis. And a word of thanks to

all faculty, staff and fellow graduate students in the Department of Mathematics who

helped and taught me in these two years.

5

Contents

ABSTRACT 2

COPYRIGHT 3

DEDICATION 4

ACKNOWLEDGEMENTS 5

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2. Multiresolutions for Type-2 Triangulations . . . . . . . . . . . . . . . 9

3. Semi-wavelets and Wavelets . . . . . . . . . . . . . . . . . . . . . . . 13

4. The Smaller Support Wavelets . . . . . . . . . . . . . . . . . . . . . . 29

4.1 Construction of the Smaller Support Wavelets . . . . . . . . . . . . . 29

4.2 Wavelet Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4.3 The Range of Parameters in the Wavelet Basis . . . . . . . . . . . . . 37

5. Parameterized Wavelet Basis . . . . . . . . . . . . . . . . . . . . . . . 41

REMARKS 54

BIBLIOGRAPHY 56

VITA 61

6

CHAPTER 1

Introduction

It is well known that wavelets have become an important tool of mathematical

analysis with a wide range of applications in mathematical physics, computational

mathematics, image compressing, detecting self-similarity in a time series and musical

tones, to mention a few (see [6,16,17,19,21]).

The first mention of wavelets appeared in an appendix to the thesis of A. Haar

(1909). One property of the Haar wavelet is that it has compact support, which

means that it vanishes outside of a finite interval. Unfortunately, Haar wavelets are

not continuously differentiable which somewhat limits their applications. In the 1930s,

several groups working independently researched the representation of functions using

scale-varying basis functions. Understanding the concepts of basis functions and scale-

varying basis functions is key to understanding wavelets.

Between 1960 and 1980, the mathematicians Guido Weiss and Ronald R. Coifman

studied the simplest elements of a function space, called atoms, with the goal of

finding the atoms for a common function and the ”assembly rules” that allow the

reconstruction of all the elements of the function space using these atoms. In 1980,

Grossman and Morlet, a physicist and an engineer, broadly defined wavelets in the

context of quantum physics. These two researchers provided a way of thinking of

wavelets based on physical intuition.

In 1985, Stephane Mallat gave wavelets an additional jump-start through his work

in digital signal processing. He discovered some relationships between quadrature

7

mirror filters, pyramid algorithms, and orthonormal wavelet bases. Inspired in part

by these results, Y. Meyer constructed the first non-trivial wavelets. Unlike the Haar

wavelets, the Meyer wavelets are continuously differentiable; however they do not

have compact support. A couple of years later, Ingrid Daubechies used Mallat’s work

to construct a set of wavelet orthonormal basis functions that are perhaps the most

elegant, and have become the cornerstone of wavelet applications today (see [5]).

Almost instantaneously it became a success story with thousands of papers written

by now with wide ranging applications. Charles K. Chui and Jianzhong Wang used

splines to construct wavelets and opened a channel to construct smooth wavelets with

short support by using spline functions (see [3]).

It is much more challenging to construct wavelets in a higher dimensional setting.

In fact, even the case of continuous piecewise linear wavelet construction is unex-

pectedly complicated (see [7,9-12,4-15]) and the references therein. However, most

real application problems are multivariate or multiparameter in nature. There creat-

ing great demand to study multivariate wavelets. In recent years, many researchers

studied wavelets over triangulations (see [7–15, 25] and references therein).

In this thesis, we emphasize the construction of piecewise linear prewavelets with

smaller support over type-2 triangulations. The thesis is organized as follows. In

Chapter 2, we introduce the concept of multiresolution over type-2 triangulations.

In Chapter 3, wavelets over type-2 triangulations are constructed by using semi-

wavelets. In Chapter 4, the smaller support wavelets are constructed. In Chapter 5,

parameterized wavelets are constructed by using the orthogonality conditions.

8

CHAPTER 2

Multiresolutions for Type-2 Triangulations

The two diagonals of each square Sij = [i, i + 1] × [j, j + 1] for i, j ∈ Z

in the plane, divide the square into four congruent triangles. Following convention,

we will refer to the set of all such triangles as a type-2 triangulation. We also refer

to any subtriangulation as a type-2 triangulation and we will be concerned with the

bounded subtriangulation τ 0, generated by the square Sij for i = 0, 1, 2, ....., m − 1

and j = 0, 1, ...., n−1, for some arbitrary m,n, see Figure 2.1. Throughout this thesis

we will assume, for the sake of simplicity, that m ≥ 2 and n ≥ 2, though wavelet

constructions can be made in a similar way when either m = 1 or n = 1 (or both).

We let V 0 and E0 denote the vertices and edges respectively in τ 0, so that

V 0 = {(i, j)}i=0,··· ,m, j=0,··· ,n ∪ {(i + 1

2, j +

1

2)}i=0,··· ,m−1, j=0,··· ,n−1

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Figure 2.2 The first refinement.

Let S0 = S01(τ

0) be the linear space of continuous functions over τ0 which

are linear over every triangle. A basis for S0 is given by the nodal functions φ0v in

S0, for v ∈ V 0, satisfying φ0v(w) = δvw. The support of φi+1/2,j+1/2 is the square

Sij , while the support of φ0ij is the diamond enclosed by the polygon with vertices

(i− 1, j),(i, j − 1),(i+ 1, j),(i, j + 1), suitably truncated if the point (i, j) lies on the

boundary of the domain D = [0, m] × [0, n].

Next consider the refined triangulation τ 1, also of type-2, formed by adding

lines in the four directions halfway between each pair of existing parallel lines, as

in Figure 2.2, and define V 1, E1, the linear space S1, and the basis φ1u, u ∈ V 1

accordingly. Then S0 is subspace of S1 and a refinement equation relates the coarse

nodal functions φ0v to the fine ones φ1

v. In order to formulate this equation we define

V 0v = {w ∈ V 0 : wand v are neighbours in V 0},

and

V 1v = {u = (w + v)/2 ∈ V 1 : w ∈ V 0

v }

Thus V 0v is the set of neighbors of v in V 1

v is the set of midpoints between v and

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its coarse neighbors. For example when v is an interior vertex, there are two cases:

V 1i+1/2,j+1/2 = {(i+1/4, j+1/4), (i+3/4, j+1/4), (i+3/4, j+3/4), (i+1/4, j+3/4)},

and

V 1i,j = {(i + 1/2, j), (i+ 1/4, j + 1/4), (i, j + 1/2), (i− 1/4, j + 1/4),

(i− 1/2, j), (i− 1/4, j − 1/4), (i, j − 1/2), (i+ 1/4, j − 1/4)}.

Then the refinement equation is easily seen to be

φ0v = φ1

v +1

2

∑u∈V 1

v

φ1u(x)

Let W 0 be the orthogonal complement space, then S1 = S0 ⊕W 0, treating S0 and

S1 as Hilbert spaces equipped with the inner product

〈f, g〉 =

∫D

f(x)g(x)dx, f, g ∈ L2(D)

Ideally we would like a basis of functions with small support for the purpose

of conveniently representing the decomposition of a given function f 1 in S1 into its

two unique components f 0 ∈ S0 and g0 ∈ W 0;

f 1 = f 0 ⊕ g0.

We will call any basis functions wavelets. Clearly the refinement of τ 0 can be

continued indefinitely, generating a nested sequence

S0 ⊂ S1 ⊂ · · ·Sk ⊂ · · · ,

and if we define the wavelet space W k−1 to be the orthogonal complement at every

refinement level k, then

Sk = Sk−1 ⊕W k−1.

11

We obtain the decomposition

Sn = S0 ⊕W 0 ⊕W 1 ⊕ · · · ⊕W n−1,

for any n ≥ 1. By combining wavelet bases for the spaces W k with the nodal bases

for the spaces Sk , we obtain the framework for a multiresolution analysis (MRA).

Note that the basis elements of any W k can simply be taken to be a dilation of the

basis elements for W 0 and therefore we restrict our study purely to W 0.

We classify these seven different structure wavelets over type-2 triangulations

into three types —- interior wavelet, boundary wavelet, and corner wavelet according

to the position of u ∈ V 1 \ V 0.

Interior wavelet: If the two neighbor vertices of u ∈ V 1 \V 0 in V 0 are the interior

vertex (not on boundary and corner), then this wavelet is called an interior wavelet.

Boundary wavelet: If at least one of the two neighbor vertices of u ∈ V 1 \ V 0 in

V 0 is the boundary vertices (not in the corner), then this wavelet is called a boundary

wavelet.

Corner wavelet: If one of the two neighbor vertices of u ∈ V 1 \ V 0 in V 0 is the

corner vertex, then this wavelet is called a corner wavelet.

12

CHAPTER 3

Semi-wavelets and Wavelets

In this chapter, we will introduce the method to construct wavelets using semi-

wavelet ideas. That is, our approach to constructing wavelets for the wavelet space

W 0 is to sum the pairs of semi-wavelets, elements of the finite space which have small

support and are close to being in the wavelet space, that they are orthogonal to all

but two of the nodal functions in the coarse space.

Letting v1 and v2 be two neighboring vertices in V 0, and denoting by u ∈ V 1 \ V 0

their midpoint, we define the semi−wavelet σv1,u ∈ S1 as the element with support

contained in the support of φ0v1

and having the property that, for all v ∈ V 0,

〈σv1,u, φ0v〉 =

−1 if v = v1;

1 if v = v2;

0 otherwise.

(3.1)

Thus σv1,u has the form

σv1,u(x) =∑

v∈Nv11

avφ1v(x)

where

Nv11= {v1} ∪ V 1

v1

denotes the fine neighborhood of v1. The only non-trivial inner products between

σv1,u and coarse nodal functions φ0v occur when v belongs to the coarse neighborhood

13

of v1,

Nv01= {v1} ∪ V 0

v1.

Thus the number of coefficients and conditions are the same and, as we will

subsequently establish, the element σv1,u is unique.

Since the dimension of W 0 is equal to the number of fine vertices in V 1, i.e.

|V 1| − |V 0|, it is natural to associate one wavelet ψu per fine vertex u ∈ V 1 \ V 0.

Since each u is the midpoint of some edge in E0 connecting two coarse vertices v1

and v2 in V 0, the element of S1,

ψu = σv1,u + σv2,u (3.2)

is a wavelet since it is orthogonal to all nodal functions φ0v, v ∈ V 0.

First we construct the interior semi-wavelets by using the definition of semi-

wavelets. Initially we consider only interior vertices v1 and there are two cases: (1)

v1 = (i+1/2, j +1/2) and (2) v1 = (i, j). Firstly, if v1 = (i+1/2, j +1/2), then σv1,u

has support contained in Sij and its fine and coarse neighborhoods are

N1v1

= {(i + 1/2, j + 1/2), (i + 1/4, j + 1/4),

(i + 3/4, j + 1/4), (i + 3/4, j + 3/4), (i+ 1/4, j + 3/4)}, (3.3)

and

N0v1

= {(i + 1/2, j + 1/2), (i, j), (i+ 1, j), (i+ 1, j + 1), (i, j + 1)} (3.4)

14

Thus there are five coefficients and five constraints imposed by the definition of

semi-wavelet and we must solve the linear system

Ax = b (3.5)

where

A = (〈φ0v, φ

1w〉)v∈N0

v1,w∈N1

v1

b =

(−1, 1, 0, 0, 0)T if v2 = (i, j);

(−1, 0, 1, 0, 0)T if v2 = (i + 1, j);

(−1, 0, 0, 1, 0)T if v2 = (i + 1, j + 1);

(−1, 0, 0, 0, 1)T if v2 = (i, j + 1);

and the ordering of the vertices in Nv01

and Nv11

is the same in (2.1) and (2.2). Due

to the symmetry, we can simply assume that b = (−1, 1, 0, 0, 0)T and the coefficients

of the remaining three semi-wavelets are the same but rotated appropriately around

v1. In order to compute the entries in the 5 × 5 matrix A. We apply the following

standard lemma.

Lemma 3.1 Let T = [x1, x2, x3] be a triangle and let f, g : T → R be two linear

functions. If fi = f(xi) and gi = g(xi) for i = 1, 2, 3, and a(T) is the area of the

triangle T, then

∫T

f(x)g(x)dx =a(T )

12(f1g1 + f2g2 + f3g3 + (f1 + f2 + f3)(g1 + g2 + g3)).

Using this lemma, and the fact that

< f, g >=∑T∈τ1

∫T

f(x)g(x)dx

15

for any f and g in S1, one can compute the entries 〈φ0v, φ

0w〉 of A and one finds that

A =1

192

20 6 6 6 63 8 1 0 13 1 8 1 03 0 1 8 13 1 0 1 8

.

Thus the vector of coefficients of σv1,u is given by

x = A−1b =1

2(−48, 64, 8, 16, 8)T

The coefficients are shown in Figure 3.1 after multiplying them by a factor of 2

(the same scaling will be applied to all later semi-wavelet coefficients). The vertex v1

is in the center of the figure (the only coarse vertex where σv1,u is non-zero) and the

fine vertex u = (v1 + v2)/2.

64

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-48

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In case (2), we suppose that v1 = (i, j), whose fine neighborhood is

N1v1

= {(i, j), (i + 1/2, j), (i+ 1/4, j + 1/4), (i, j + 1/2), (i− 1/4, j + 1/4),

(i− 1/2, j), (i− 1/4, j − 1/4), (i, j − 1/2), (i+ 1/4, j − 1/4)},

16

and whose coarse neighborhood is

N0v1

= {(i, j), (i+ 1, j), (i + 1/2, j + 1/2), (i, j + 1), (i− 1/2, j + 1/2),

(i− 1, j), (i− 1/2, j − 1/2), (i, j − 1), (i + 1/2, j − 1/2)},

Thus we again solve the linear system (3.5) where A is this time a 9 × 9 matrix

and b is either

(−1, 1, 0, 0, 0, 0, 0, 0)T or (−1, 0, 1, 0, 0, 0, 0, 0)T

depending on whether v2 = (i + 1/2, j) or v2 = (i + 1/4, j + 1/4) and the remaining

six possible coarse neighbors v2 lead to the same coefficients , only rotated. Applying

Lemma 3.1 we find after some calculation that

A =1

192

24 12 8 12 8 12 8 12 81 12 1 0 0 0 0 0 11 4 6 4 0 0 0 0 01 0 1 12 1 0 0 0 01 0 0 4 6 4 0 0 01 0 0 0 1 12 1 0 01 0 0 0 0 4 6 4 01 0 0 0 0 0 1 12 11 4 0 0 0 0 0 4 6

.

Thus if b = (−1, 1, 0, 0, 0, 0, 0, 0, 0)T , we find that

x = A−1b =1

2(−24, 38,−24, 4, 0, 2, 0, 4,−24)T ,

and if b = (−1, 0, 1, 0, 0, 0, 0, 0, 0)T , we find that

x = A−1b =1

2(−48,−3, 76,−3, 8, 3, 4, 3, 8)T .

These two semi-wavelets are illustrated in Figures 3.2 and 3.3. Using the three

interior semi-wavelets of Figure 3.1, 3.2 and 3.3 provides us with two wavelets ψu,

17

from (3.2). The first of these, in Figure 3.4, is the sum of two semi-wavelets from 3.2

and the second, in Figure 3.4, the sum of the semi-wavelets in Figures 3.1 and 3.3.

Symmetries and rotations of these two give us all interior wavelets ψu in the sense

that v1 and v2 are both interior vertices of τ 0.

2

0

0

4

-24

4

-24

-24

38✍✌✎�

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❅❅

Figure 3.2. Second interior semi-wavelet.

3

4

8

3

-48

3

8

76

-3

✍✌✎

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❅❅

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❅❅

Figure 3.3. Third interior semi-wavelet.

18

2

0

0

4

-24

4

-24

-24

76

-24

-24

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-24

4

0

0

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Figure 3.4. First interior wavelet.

3

4

8

3

-48

-3

8

140

8

-3

-48

8

16

✖✕✗✔

✉

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Figure 3.5. Second interior wavelet.

Now we consider the case where v1 is a boundary vertex, which means that v1 =

(i, j). Let us suppose first that v1 lies on an edge of the domain, but not at one of the

four corners, thus we assume without loss of generality that j = 0 and 0 < i < m.

The coarse and fine neighborhoods of v1 are then

N1v1

= {(i, 0), (i + 1, 0), (i+ 1/2, 1/2), (i, 1), (i− 1/2, 1/2), (i− 1, 0),

19

and

N0v1

= {(i, 0), (i+ 1/2, 0), (i+ 1/4, j + 1/4), (i, 1/2), (i− 1/4, 1/4), (i− 1/2, 0),

respectively, and the matrix A has dimension 6 × 6. From Lemma 3.1 we find that

A =1

192

12 6 8 12 8 61/2 6 1 0 0 01 4 6 4 0 01 0 1 12 1 01 0 0 4 6 4

1/2 0 0 0 1 6

.

If v2 = (i+1, j), then b = (−1, 1, 0, 0, 0, 0)T and x = A−1b = 12(−48, 76,−48, 8, 0, 4)T .

If v2 = (i+1/2, j+1/2), then b = (−1, 0, 1, 0, 0, 0)T and x = A−1b = 12(−96,−6, 84, 0, 12, 6)T .

If v2 = (i, j+1) then b = (−1, 0, 0, 1, 0, 0)T and x = A−1b = 12(−48, 8,−24, 40,−24, 8)T .

These three elements are shown in Figures 3.6, 3.7, and 3.8. Summing two of the

first boundary semi-wavelets gives us the boundary wavelet ψu shown in Figure 3.9.

Summing the second boundary semi-wavelet and the first interior semi-wavelet gives

us the wavelet ψu shown in Figure 3.10. Finally, summing the third edge semi-wavelet

and the second interior semi-wavelet gives us the edge wavelet ψu shown in Figure

3.11. Up to rotation and symmetries these elements provide all wavelets ψu for which

one of v1 and v2 is an interior vertex while the other one lies on the boundary but

not at a corner.

4

0

-48

8

-48

76✖✕✗✔✉

✉

✉

✉

✉❡

❡ ❡

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❅❅

❅❅❅

❅❅

❅❅❅

❅❅

❅❅

Figure 3.6. First boundary semi-wavelet.20

6

12

-96

0

84

-6

✖✕✗✔

✉

✉

✉

✉

✉❡

❡ ❡

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❅❅

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❅❅

Figure 3.7. Second boundary semi-wavelet.

8

-24

-48

40

-24

8

✖✕✗✔

✉

✉

✉

✉

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Figure 3.8. Third boundary semi-wavelet.

4

0

-48

8

-48

152

-48

-48

8

0

4✖✕✗✔✉

✉

✉

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��

Figure 3.9. First boundary wavelet

21

6

12

-96

0

148

8

-6

-48

8

16

✖✕✗✔

✉

✉

✉ ❡

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��

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❅❅❅

Figure 3.10. Second boundary wavelet.

8

-24

-24

-24

0

-48

78

-24

2

-24

-24

0

8

4

✖✕✗✔

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❅❅

Figure 3.11. Third boundary wavelet.

In the case that v1 is one of the four corners of the domain, we may suppose

without loss of generality that v1 = (0, 0). The coarse and fine neighborhoods of v1

are then

N0v1

= {(0, 0), (1, 0), (1/2, 1/2), (0, 1)},

22

and

N0v1

= {(0, 0), (1/2, 0), (1/4, 1/4), (0, 1/2)},

and the matrix A has dimension 4× 4, specifically,

A =1

192

6 6 8 61/2 6 1 01 4 6 4

1/2 0 1 6

.

There are only two cases, up to symmetry: if v2 = (1, 0) then b = (−1, 1, 0, 0)T

and x = A−1b = 12(−96, 80,−48, 16)T ; while if v2 = (1/2, 1/2) then b = (−1, 0, 1, 0)T

and x = A−1b = 12(−192, 0, 96, 0)T ; the two semi-wavelets are shown in Figures 3.12,

and 3.13. Summing the first corner semi-wavelet and the first boundary semi-wavelet

yields the wavelet in Figure 3.14 and summing the second corner semi-wavelet and the

first interior semi-wavelet yields the wavelet in Figure 3.15. Symmetries and rotations

of these give us all remaining wavelets ψu.

-96

16

-48

80✖✕✗✔

✉

✉

✉

❡

❡

❅❅

❅❅

❅❅

❅❅

❅❅

❅❅

��

��

��

��

❅❅

❅❅

Figure 3.12. First corner semi-wavelet.

23

-192

0

96

0

✖✕✗✔

✉

✉

✉

❡

❡

❅❅

❅❅

❅❅

❅❅

❅❅

❅❅

��

��

��

��

❅❅

❅❅

Figure 3.13. Second corner semi-wavelet.

-96

16

-48

156

-48

-48

8

0

4✖✕✗✔

✉

✉

✉

✉

✉

❡ ❡ ❡

❡�

�

��

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❅❅

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❅❅

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❅❅

❅❅

❅❅❅

❅❅❅

❅❅

❅❅

��

Figure 3.14. First corner wavelet.

-192

0

160

8

0

-48

8

16

✖✕✗✔

✉ ❡

✉

❡

✉

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❅❅

❅❅

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❅❅❅�

�

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��

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❅❅

❅❅❅

Figure 3.15. Second corner wavelet.

In order to prove that the above wavelets consist of wavelet basis, we introduce the

following Lemma.

Lemma 3.2 A set of wavelets Ψ = (ψu1 , ψu2 , · · · , ψun) in W 0 is a basis of W 0 if the

matrix Q = (qui,uj)i,j is nonsingular, where qui,uj

= ψuj(ui).

24

Proof. Given a linear combination∑n

j=1 cjψuj, which is identically zero, evaluation

at ui yieldsn∑

j=1

cjψuj(ui) =

n∑j=1

cjqui,uj= 0

and so Qc = 0 where c = (c1, c2, c3, · · · , cn)T , Therefore c = 0.

In the following, we will prove that the wavelets defined by (3.2) can consist of

the wavelet basis.

Theorem 3.3 The set of wavelets {φu}u∈V 1\V 0 defined by (3.2) is a basis for the

wavelet space W 0.

Proof. It is sufficient to show that the wavelets φu are linearly independent. We

demonstrate this by showing that the square matrix

Q = (φv(u))u,v∈V 1\V 0

is diagonally dominant and therefore non-singular. Diagonal dominance is clearly

equivalent to the condition that

φv(v) >∑

u∈V 1\V 0,u �=v

|φu(v)|, for all v ∈ V 1 \ V 0.

Thus for each v in V 1\V 0 we need to show the sum of the absolute values of coefficients

at v of wavelets other than φv is less than the coefficient at v of φv itself. It turns out

that this condition does indeed hold in every topological case. In Figure 3.16 each

distinct topological case of v ∈ V 1 \ V 0 is illustrated by placing the value φu(v) at u

for each relevant u. The vertex v is circled in each case. Thus the coefficients in each

figure are the non-zero elements of the row v of the matrix Q.

25

2

3

3

4

4

-3

-3

76

-3

-3

4

4

3

3

2✉ ✉

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Figure 3.16. Wavelets evaluations

28

CHAPTER 4

The Smaller Support Wavelets

In Chapter 3, we used the semi-wavelet idea to obtain the small support and

unique wavelets over type-2 triangulations. In this chapter, we will work on second

interior wavelet and second boundary wavelet—the smaller support interior wavelet

and boundary wavelet will be constructed.

4.1 Construction of the Smaller Support Wavelets

First, we will work on the second boundary wavelet. Support vertices are labeled

in the following Figure 4.1 and Pi i = 1, · · · , 6 are labeled for the old vertices. This

wavelet is called the second smaller support boundary wavelet.

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4

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u

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P2P3

P4

P5

Figure 4.1. The second smaller support boundary wavelet

Let φ0u be the wavelet function at u which has the following expression:

φ0u = Aφ1

u + B1φ11 + B2φ

12 + B3φ

13 + B4φ

14 + B5φ

15 + B6φ

16 + B7φ

17

Where φ0u is a wavelet function at vertex u in W 0, and φ1

i , i = u, 1, · · · , 7 are nodal

basis functions at u, i = 1, · · · , 7 in S1. By the orthogonal conditions, the following

29

inner products must be zeros,

〈φ0u, φ

01〉 = 0, 〈φ0

u, φ0pi〉 = 0, i = 1, · · ·5, 〈φ0

u, φ06〉 = 0

By Lemma 3.1 and computation, we obtain the following linear equations:

8A + 12B1 + 6B2 + 12B3 + 8B4 + 6B5 + 3B6 + 0B7 = 0,

1A + 12B1 + 6B2 + 0B3 + 0B4 + 0B5 + 3B6 + 1B7 = 0,

0A + 0B1 + 0B2 + 0B3 + 0B4 + 0B5 + 3B6 + 8B7 = 0,

1A + 1B1 + 0B2 + 12B3 + 1B4 + 0B5 + 3B6 + 1B7 = 0,

0A + 1B1 + 0B2 + 4B3 + 6B4 + 4B5 + 0B6 + 0B7 = 0,

0A + 12B1 + 0B2 + 0B3 + 1B4 + 6B5 + 0B6 + 0B7 = 0,

6A + 1B1 + 4B2 + 4B3 + 0B4 + 0B5 + 20B6 + 6B7 = 0,

The coefficient matrix of the above linear equations is

D1 =

8 12 6 12 8 6 3 01 1

26 0 0 0 3 1

0 0 0 0 0 0 3 81 1 0 12 1 0 3 10 1 0 4 6 4 0 00 1

20 0 1 6 0 0

6 1 4 4 0 0 20 6

.

Let the vector v1 = [A,B1, B2, B3, B4, B5, B6, B7]T , the solutions of

D1v1 = 0

be,

v1 =1

2

[204

5k,−144

5k,

6

5k,

8

5k,

12

5k, 2k,−64

5k,

24

5k

]T

, (1)

30

where k is a non-zero arbitrary real number.

From the above solutions, it is clear that the second smaller support boundary

wavelet only needs 8 points of support but second boundary wavelet needs 9 points

of support and is not unique, depending on the value of k.

In the following, we study the second smaller support interior wavelet based on

the same structure as second interior wavelet. The support vertices are labeled and

some vertices in V 0 are also labeled in the following Figure 4.2. This wavelet is called

the second smaller support interior wavelet.

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P1

P2P3

P4

P5

P6

P7

P8

Let’s assume that the wavelet at u has the following expression:

φ0u = B1φ

11 + Aφ1

u + B4φ14 + B5φ

15 + B6φ

16

+B7φ17 + B8φ

18 + B9φ

19 + B10φ

110 + B11φ

111 + B12φ

112

31

Where φ1u is a basis wavelet function in W 0, and φ1

i , i = u, 1, · · · , 12 are nodal basis

functions in S1.

By the orthogonal conditions, 〈φ0u, φ

0Pi〉 = 0, i = 1, · · · , 8 and 〈φ0

u, φ01〉 = 0,

〈φ0u, φ

010〉 = 0, we will obtain the following linear equations:

24B1 + 8A + 8B4 + 12B5 + 8B6 + 12B7 + 8B8 + B9 + 3B10 + 0B11 + B12 = 0

B1 + 1A + 0B4 + 0B5 + 0B6 + 0B7 + 1B8 + 8B9 + 3B10 + 1B11 + 0B12 = 0

B1 + 6A + 0B4 + 0B5 + 0B6 + 0B7 + 0B8 + 6B9 + 20B10 + 6B11 + 6B12 = 0

B1 + 1A + 1B4 + 0B5 + 0B6 + 0B7 + 0B8 + 0B9 + 3B10 + B11 + 8B12 = 0

B1 + 0A + 6B4 + 4B5 + 0B6 + 0B7 + 0B8 + 0B9 + 0B10 + 0B11 + 0B12 = 0

B1 + 0A + B4 + 12B5 + 1B6 + 0B7 + 0B8 + 0B9 + 0B10 + 0B11 + 0B12 = 0

B1 + 0A + 0B4 + 4B5 + 6B6 + 4B7 + 0B8 + 0B9 + 0B10 + 0B11 + 0B12 = 0

B1 + 0A + 0B4 + 0B5 + B6 + 12B7 + B8 + 0B9 + 0B10 + 0B11 + 0B12 = 0

B1 + 0A + 0B4 + 0B5 + 0B6 + 4B7 + 6B8 + 4B9 + 0B10 + 0B11 + 0B12 = 0

0B1 + 0A + 0B4 + 0B5 + 0B6 + 0B7 + 0B8 + B9 + 3B10 + 8B11 + B12 = 0

32

The coefficient matrix of the above linear equations is the following:

D2 =

24 12 8 12 8 12 8 12 8 1 3 0 11 12 1 0 0 0 0 0 1 8 3 1 01 4 6 4 0 0 0 0 0 6 20 6 61 0 1 12 1 0 0 0 0 0 3 1 81 0 0 4 6 4 0 0 0 0 0 0 01 0 0 0 1 12 1 0 0 0 0 0 01 0 0 0 0 4 6 4 0 0 0 0 01 0 0 0 0 0 1 12 1 0 0 0 01 4 0 0 0 0 0 4 6 0 0 0 00 0 0 0 0 0 0 0 0 1 3 8 1

.

We solve this linear equation and obtain the following solutions

B1 = −15k1, A =253

6k1, B4 =

11

6k1, B5 = k1, B6 =

7

6k1, B7 = k1

B8 =11

6k1, B9 = k1, B10 = −14k1, B11 = 5k1, B12 = k1

where k1 is a non-zero arbitrary real number.

This second smaller support interior wavelet only needs 11 points of support, but

second interior wavelet needs 13 points of support.

Due to symmetry of type-2 triangulations, rotation will generate all types of

wavelets which have the same structures as these two wavelets.

4.2 Wavelet Basis

Theorem 4.4 Let k = 5 and k1 = 3. Then the set of {φ0u}u∈V 1\V 0 which contains the

first interior wavelet, the second smaller support interior wavelet, the first boundary

wavelet, the second smaller support boundary wavelet, the third boundary wavelet, the

first corner wavelet and the second corner wavelet is a basis for wavelet space W0.

33

Proof. It is sufficient to show that the wavelets {φ0u}u∈V 1\V 0 are linearly independent.

We demonstrate this by showing that the following square matrix

Q = (φ0v(u))u,v∈V1\V0

is diagonally dominant and therefore non-singular. Diagonal dominance is clearly

equivalent to the condition that

|φv(v)| >∑

u∈V1\V0,u �=v

|φu(v)| mboxfor all v ∈ V1 \ V0

By computing the value of every wavelet at u, we can get non-zero values in each

row of Q as shown in the Figure 4.3, and it is sufficient to prove that if every row

is dominant in matrix Q, then Q is non-singular matrix, that is, these wavelets can

consist of wavelet basis.

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From the above discussion, we know that we can choose different k,k1 to ensure

that these two smaller support wavelets combining other five structure wavelets in

Chapter 3 can consist of wavelet basis for wavelet space W 0.

36

4.3 The Range of Parameters in the Wavelet Basis

In this section, we will give the sufficient conditions of k and k1 to ensure that the

first interior wavelet, the second smaller support interior wavelet, the first boundary

wavelet, the second smaller support boundary wavelet, the third boundary wavelet,

the first corner wavelet, and the second corner wavelet can consist of wavelet basis

for wavelet space W 0.

Theorem 4.5 If k and k1 satisfy the following conditions,

144

149< |k| < 34

5

2645

1192< |k1| < 20

4|k| + 5|k1| < 65

Then the set of {φ0u}u∈V 1\V 0 which contains the first interior wavelet, the sec-

ond smaller support interior wavelet, the first boundary wavelet,the second boundary

wavelet, the third boundary wavelet, the first corner wavelet and the second corner

wavelet is a basis for wavelet space W0.

Proof. We consider the first interior wavelet, the second smaller support interior

wavelet, the first boundary wavelet, the second smaller support boundary wavelet,

the third boundary wavelet, the first corner wavelet and the second corner wavelet.

In order to verify that these wavelets can consist of the basis, we need to prove that

the following matrix is non-singular

Q = (φ0u(v))v∈V 1\V 0.

37

Figures 4.4 shows every non-zero values in every row of Q. We obtain the following

dominant inequalities in every row from each figure.

253

3|k1| > 48 +

11

3|k1| + 22

3|k1| + 10|k1|

204

5|k| > 48 + 24 +

12

5|k| + 22

6|K| + 10|K|

76 > 8|k1| + 8 + 8 + 4

102 > 24 + 4|k1| + 12

5|k|

78 > 16 + 8 + 2 + 4|k1| + 16

5|k|

156 > 28 +16

5|k|

160 > 92 + 10|k1|.

We solve the above inequalities— the range for k and k1 in the theorem will be

obtained.

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4

k

8

65k

152

k

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0

125

k

-24

2045

k

113

k1

73k1

-48

0

73k1

10k1

✉

✉

✖✕✗✔

✉

✉

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✉

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❡

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39

8

4

85k

0

2k1

78

2

85k

0

2k1

8

4

✉

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✉

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0

156

65k

8

2k

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-48

160

0

-48

0

10k1

73k1

✉

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�

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��❅

❅❅

❅❅❅

Figure 4.4. Evaluation of wavelets

40

CHAPTER 5

Parameterized Wavelet Basis

In this chapter, we will construct the parameterized wavelet basis over type-2

triangulations. The two smaller support wavelets are discussed in Chapter 4 (See the

Figures 5.1 and 5.2). Since there are parameters in these two wavelets, we call the

following wavelets parameterized wavelet 1 and parameterized wavelet 2, respectively.

t1

76t1

116

t1

t1

-15t1

116

t1

2536

t1

t1

-14t1

t1

5t1

❡

❡✖✕✗✔

✉

✉

✉

✉

✉

✉

✉

✉

❡

❡

❡

❡

❡

❡

❡

❡

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��

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��❅❅❅

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��

❅❅❅

❅❅

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❅❅

❅❅❅

❅❅❅

❅❅

❅❅

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❅❅❅

��

��

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Figure 5.1. Parameterized Wavelet 1

t2

65t2

- 725

t2

45t2

1025

t2

35t2

- 325

t2

125

t2❡

❡✖✕✗✔

✉

✉

✉ ❡

✉

✉

❡

❡❡

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��

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��

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❅❅❅

❅❅❅


41

In fact, the above wavelets are interior and boundary wavelets. According to

symmetry, the rotation of the two wavelets can form the other interior and edge

wavelets which have the same structure as these two parameterized wavelets.

We will construct the other five paramaterized wavelets directly. First we consider

the following figure and label the vertices in the Figure 5.3.

5

6

4

7

1

3

8

2

u

9

16

10

13

15

11

14

12

�

✍✌✎�

�

�

�

�

�

�

�

�

�

�❝

❝

❝

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�

❅❅

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❅❅

❅

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��

P1

P2

P12

P3

P11

P4

P10

P5

P9

P6

P7

P8


Let σu be the wavelet at vertex u with the following expression:

σu = Aφ1u + B1φ

11 + B2φ

12 + B3φ

13 + B4φ

14 + B5φ

15 + B6φ

16 + B7φ

17 + B8φ

18

+B9φ19 + B10φ

110 + B11φ

111 + B12φ

112 + B13φ

113 + B14φ

114 + B15φ

115 + B16φ

116

Here A and Bi (i = 1, · · · , 16) will be determined by using the orthogonality

conditions. By using the orthogonal conditions such as 〈σu, φ0Pi〉 = 0, i = 1, · · · , 12

and 〈σu, φ01〉 = 0,〈σu, φ

013〉 = 0, we obtain the following equations:

42

4A + B1 + 6B2 + 4B3 = 0

B1 + B2 + 12B3 + B4 = 0

B1 + 4B3 + 6B4 + 4B5 = 0

B1 + B4 + 12B5 + B6 = 0

B1 + 4B5 + 6B6 + 4B7 = 0

B1 + B6 + 12B7 + B8 = 0

4A + 4B7 + 6B8 + 6B9 + 4B10 = 0

B9 + 12B10 + B11 + B13 = 0

4B10 + 6B11 + 4B12 + B13 = 0

B11 + 12B12 + B13 + B14 = 0

4B12 + B13 + 6B14 + 4B15 = 0

12A + B1 + B3 + B8 + 8B9 + 12B10+

8B11 + 12B12 + 24B13 + 8B14 + 12B15 + 8B16 = 0

Let the vector v1 = [A,B1, B2, · · · , B16]T , then the coefficient matrix of the above

linear equation is

43

M1 =

4 1 6 4 0 0 0 0 0 0 0 0 0 1 0 4 60 1 1 12 1 0 0 0 0 0 0 0 0 0 0 0 00 1 0 4 6 4 0 0 0 0 0 0 0 0 0 0 00 1 0 0 1 12 1 0 0 0 0 0 0 0 0 0 00 1 0 0 0 4 6 4 0 0 0 0 0 0 0 0 00 1 0 0 0 0 1 12 1 0 0 0 0 0 0 0 04 1 0 0 0 0 0 4 6 6 4 0 0 1 0 0 00 0 0 0 0 0 0 0 0 1 12 1 0 1 0 0 00 0 0 0 0 0 0 0 0 0 4 6 4 1 0 0 00 0 0 0 0 0 0 0 0 0 0 1 12 1 1 0 00 0 0 0 0 0 0 0 0 0 0 0 4 1 6 4 00 0 0 0 0 0 0 0 0 0 0 0 0 1 1 12 112 24 8 12 8 12 8 12 8 1 0 0 0 1 0 0 112 1 1 0 0 0 0 0 1 8 12 8 12 24 8 12 8

.

We let B14 = 0, B11 = 0,B6 = 0,and B4 = 0, and the coefficient matrix will be

transformed into the following submatrix:

M11 =

4 1 6 4 0 0 0 0 0 0 1 4 60 1 1 12 0 0 0 0 0 0 0 0 00 1 0 4 4 0 0 0 0 0 0 0 00 1 0 0 12 0 0 0 0 0 0 0 00 1 0 0 4 4 0 0 0 0 0 0 00 1 0 0 0 12 1 0 0 0 0 0 04 1 0 0 0 4 6 6 4 0 1 0 00 0 0 0 0 0 0 1 12 0 1 0 00 0 0 0 0 0 0 0 4 4 1 0 00 0 0 0 0 0 0 0 0 12 1 0 00 0 0 0 0 0 0 0 0 4 1 4 00 0 0 0 0 0 0 0 0 0 1 12 112 24 8 12 12 12 8 1 0 0 1 0 112 1 1 0 0 0 1 8 12 12 24 12 8

.

We solve the following equation M1v1 = 0, where 0 is a column vector.

v1 = [38t3,−12t3,−12t3, 2t3, 0, t3, 0, 2t3,−12t3,−12t3,−12t3, 0, 2t3, t3, 0,−12t3, 2t3,−12t3]T

where t3 is an arbitrary non-zero real number.

44

By the similar way, we label the support vertices and compute the coefficients

over the Figure 5.4.

5

4

1

3

2

u

10

6

9

8

7✖✕✗✔✉

✉

✉

✉

✉

✉

✉❡

❡ ❡ ❡ ❡

❡�

��

��

��

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❅❅

❅❅

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❅❅

❅❅�

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❅❅

❅❅

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❅❅

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��

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��

��

❅❅

❅❅

❅❅❅

❅❅❅

❅❅

��

��

❅❅❅

��


First, we assume that the wavelet at vertex u has the following expression:

σu = Aφ1u + B1φ

11 + B2φ

12 + B3φ

13 + B4φ

14 + B5φ

15+

B6φ16 + B7φ

17 + B8φ

18 + B9φ

19 + B10φ

110.

By the orthogonality conditions, we will obtain the matrix:

M2 =

6 12 8 12 8 6 1/2 0 0 0 16 1/2 1 0 0 0 12 6 8 12 80 0 0 0 0 0 1/2 6 1 0 00 0 0 0 0 0 1 4 8 4 00 0 0 0 0 0 1 0 1 12 14 1 6 4 0 0 1 0 0 4 60 1 1 12 1 0 0 0 0 0 00 1 0 4 6 4 0 0 0 0 00 1/2 0 0 1 6 0 0 0 0 0

.

We let B4 = 0 and B8 = 0, and obtain the coefficient matrix:

45

M22 =

6 12 8 12 6 1/2 0 0 16 1/2 1 0 0 12 6 12 80 0 0 0 0 1/2 6 0 00 0 0 0 0 1 4 4 00 0 0 0 0 1 0 12 14 1 6 4 0 1 0 4 60 1 1 12 0 0 0 0 00 1 0 4 4 0 0 0 00 1/2 0 0 6 0 0 0 0

.

Let the column vector v2 = [A,B1, B2, B3, B5, B6, B7, B8, B9, B10]T be the column

vector. By solving the linear equation M22v2 = 0, where 0 is a column vector, we will

obtain the solutions

v2 = [38t4,−12t4,−12t4, 2t4, t4,−12t4, t4, 2t4,−12t4]T


5

11

4

12

10

1

u

6

9

3

13

8

2

7

✖✕✗✔

✉

✉

✉

✉

✉

✉

✉

✉

✉

❡

❡

❡

❡ ❡

❡

❡

❡

��

��

��

��

��

��

��

❅❅

❅❅

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❅❅❅

❅❅❅ �

�

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��

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��

��

❅❅

❅❅

❅❅

❅❅

❅❅❅

❅❅❅

❅❅

❅❅

❅❅❅❅

❅❅

❅❅

❅❅

❅❅❅�

��

��

��

��

❅❅❅

❅❅


We consider the boundary wavelet at u in the Figure 5.5 and it has the expression:

σu = Aφ1u + B1φ

11 + B2φ

12 + B3φ

13 + B4φ

14 + B5φ

15 + B6φ

16+

46

B7φ17 + B8φ

18 + B9φ

19 + B10φ

110 + B11φ

111 + B12φ

112 + B13φ

113.

By the orthogonality conditions, we obtain the following linear equations:

12A + B1 + B3 + B4 + 24B6 + 12B7 + 8B8

+12B9 + 8B10 + 12B11 + 8B12 + 8B13 = 0

12A + 12B1 + 6B2 + 8B3 + 8B4 + 6B5 + B6 + B13 + B14 = 0

12B1 + 6B2 + B3 = 0

4A + B1 + 4B2 + 6B3 + B6 + 4B7 + 6B13 = 0

B6 + 12B7 + B8 + B13 = 0

B6 + 4B7 + 6B8 + 4B9 = 0

B6 + B8 + 12B9 + B10 = 0

B6 + 4B9 + 6B10 + 4B11 = 0

B6 + B10 + 12B11 + B12 = 0

4A + B1 + 6B4 + 4B5 + B6 + 4B11 + 6B12 = 0

12B1 + B4 + 6B5 = 0

By letting B10 = 0 and B8 = 0, and letting the vector v3 be the following form:

v3 = [A,B1, B2, B3, B4, B5, B6, B7, B9, B11, B12, B13]T ,

we solve the deleted linear equations and obtain the following solutions:

v3 = [39t5,−24t5, 4t5,−12t5,−12t5, 4t5,−12t5, 2t5, 0, t5, 0, 2t5,−12t5,−12t5]T

47


Let’s consider the other structure wavelet in the Figure 5.6 and label the vertices

in the Figure 5.6,

1

3

2

u

5

4

8

7

6✖✕✗✔

✉

✉

✉

✉

✉

❡ ❡ ❡

❡�

�

��

��

��

��❅

❅❅

❅❅

❅❅

❅❅

❅❅❅

❅❅

❅❅

❅❅

❅❅❅

❅❅��

��

��

��

��

��

❅❅

❅❅ ❅❅

❅❅

❅❅

❅❅

❅❅❅

❅❅❅

❅❅

❅❅

��


Let σu be the wavelet on the vertex u and suppose it has the following expression:

σu = Aφ1u + B1φ

11 + B2φ

12 + B3φ

13 + B4φ

14 + B5φ

15 + B6φ

16 + B7φ

17 + B8φ

18.

Assume that vector v4 has the following expression:

v4 = [A,B1, B2, B3, B4, B5, B6, B7, B8]T .

By the orthogonality conditions and letting B7 = 0, we obtain the coefficient matrix

M4 of linear equation M4v4 = 0:

M4 =

6 6 8 6 1/2 1 0 06 1/2 1 0 12 8 6 120 0 0 0 1/2 0 6 00 0 0 0 1 0 4 40 0 0 0 1 1 0 124 1 6 4 1 6 0 40 1/2 1 6 0 0 0 0

.

48

Solving the linear equation M4v4 = 0, we obtain the following solution:

v5 = [39t6,−24t6,−12t6, 4t6,−12t6,−12t6, t6, 0, 2t6]T .


Let’s consider the seventh structure wavelet in the following figure and label the

vertices on the Figure 5.7.

1

3

u

7

2

5

4

6

✖✕✗✔

✉ ❡

✉

❡

✉

��

��

��

��❅

❅❅

❅❅

❅❅

❅❅❅�

�

��

❅❅

❅❅

��

��❅

❅❅

❅❅❅


Let σu be the wavelet on vertex u in the parameterized wavelet 7 and suppose it

has the expression:

σu = Aφ1u + B1φ

11 + B2φ

12 + B3φ

13 + B4φ

14 + B5φ

15 + B6φ

16 + B7φ7.

Assume that vector v5 has the expression:

v5 = [A,B1, B4, B5, B6, B7]T .

Let B2 = 0 and B3 = 0 ,by the orthogonality conditions, we obtain the coefficient

matrix M5 of linear equation M5v5 = 0:

M5 =

6 1 6 20 6 68 6 1 3 0 11 1/2 8 3 1 00 0 1 3 8 11 1/2 0 3 1 8

.

49

By solving the linear equation M5v5 = 0, we obtain the solution:

v5 = [20t7,−24t7, t7,−6t7, 2t7, t7]T


Due to the symmetry, the rotations of the above seven parameterized wavelets

can form any wavelet functions on the new vertex in u ∈ V 1 \ V 0. So we can obtain

all wavelet functions in W 0.

When t3 = 1, t4 = 1, t5 = 2, t6 = 2, and t7 = 4, the above five wavelets can

be transformed into the first interior wavelet, the first boundary wavelet, the third

boundary wavelet, the first corner wavelet, and the second corner wavelet in Chapter

3. These parameterized wavelets have smaller support and are not unique depending

on the parameters ti. In the following, we will give sufficient conditions of these

parameters ti, i = 1, · · · , 7 to ensure that these seven wavelets can form a wavelet

basis.

Theorem 5.6 We consider the above seven parameterized wavelets 1, · · · 7, if ti,i =

1, · · · , 7 in the parameterized wavelets 1, · · · , 7 satisfy the following conditions,

144

149|t3| < |t1| < min{7|t3|, 5|t7 − 6|t6|}

5

96(41

6|t1| + 12|t4| + 12|t5|) < |t2| < min(

5

8(18|t4 − 4|t5|),

5

8(39|t5| − 4|t4| − 5|t3| − 2|t1|, 1

2(35|t6| − 4|t5| − 5|t4|))

where ti �= 0.

Then these seven parameterized wavelets can consist of a wavelet basis.

50

Proof. Let Q = (φ1u(v))v∈V 1\V 0 be a matrix evaluated at u by every parameterized

wavelet. The following figures show that the non-zero values of rows in matrix Q.

t3

t1

t1

2t3

2t3

0

0

38t3

0

0

2t3

2t3

t1

t1

t3✉ ✉✖✕✗✔✉

✉

✉

✉

✉

✉

✉

✉

✉

✉

✉❡

❡

❡

❡

❡

❡

❡

❡

❡

❡

❡

❡

❅❅

❅❅

❅❅

❅❅

❅❅

❅❅

��

��

��

��

��

��

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��

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��

��

��

��

❅❅

❅❅

❅❅

❅❅❅

❅❅

❅❅

❅❅❅

��

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52

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Figure 5.8 Evaluation of parameterized wavelets

It is easy to verity that if the ti satisfy the above conditions, then matrix Q is

a row dominant matrix, and Q is nonsingular, that is, these parameterized wavelets

can consist of the wavelet basis for wavelet space W 0 .

It is clear that we can construct the smaller support wavelets over type-2

triangulations. These smaller support wavelets combining other wavelets can consist

of the wavelet basis on the wavelet space. Parameterized wavelets are proposed

and constructed by using parameters and these parameterized wavelets can form

53

the wavelet basis when these parameters satisfy some conditions, that is, we can pick

infinite ti i = 1, · · · , 7 to obtain the wavelet basis.

54

Remarks

1: In Chapter 4, we construct the smaller support wavelets over a type-2 triangu-

lation. Does the smallest support linear piecewise wavelet basis exist over the type-2

triangulations?

2: Does the smallest support linear piecewise parameterized wavelet basis exist

over this type-2 triangulation?

3: In Chapter 5, we prove that parameterized wavelets can consist of a wavelet

basis when parameters satisfy some conditions. Is it possible to find the better bounds

for ti to ensure that these seven parameterized wavelets can form a wavelet basis?

55

BIBLIOGRAPHY

56

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1482.

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[4] Chui, C.K., Multivariate Splines, Philadephia, 1988.

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[10] Floater, M. S.and Quak, E. G., Linear Independence and Stability of Piece-

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(2000), 58–79.

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angulations, Numer. Math. 82 (1999), 221–252.

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Press, Nashville, (1998), 60-73.

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60

VITA

Jiansheng Cao

Education: Shandong Technology University, Shandong, P. R. China;Mathematics, B.S. 1986.

East Tennessee State University, Johnson City, Tennessee;

Mathematics, M.S. 2002.

Experience: ETSU, August, 2000-August, 2002;Graduate Assistant, East Tennessee State University;

Johnson City, Tennessee, Sept. 2000 – Aug. 2002

Publications: With Dr. Gardner, Restrictions on the Zeros of a Polynomialas a Consequence of Conditions on the Coefficients of

Even Powers and Odd Powers of the Variable, Submitted.With Dr. Hong, Best Approximation for Symmetric Semi-definite

Positive Solutions of the Left and Right InverseEigenavalue Problems on a Subspace, Presentation in

970th AMS MEETING Oct.5-6,2001 at Chattanooga, Tennessee.The Two Classes of Best Approximation of a Matrix on the Linear

Manifold,Mathematica Numerica Sinica 20(2)1998, 148-152.

The Positive Definite and Semi-definite Solutions of AX = B on Subspace,Journal of Engineering Mathematics, Vol.(17) No.3, 2000, 31-35.

A Class of the Best Approximation of Matrices Under Restriction, AppliedMathematics- A Journal of Chinese universities,11(2)1996, 219-224.

Generalized Inverse Eigenivalue Problem of Matrix and Its Applications,Proceedings of 18th International Conference On Computer

and Industrial Engineering, China Machine Press, 10/95.

61

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Construction of Piecewise Linear Wavelets